Lab 7 Data

Part 3: Elastic Collisions Slide 5: Immovable Object Remote, m 1 = 458 g =0.458 kg Wall, m 2 = EXTREMELY BIG V i remote = -0.0205 m/s V f remote = 0.0205 m/s ΔKE = ½ (0.458 kg)(0.0205 m/s) 2 – ½ (0.458 kg)(-0.0205 m/s) 2 = 0.0 J The data shows the remote having an initial velocity of - 0.205 m/s, accelerating to a peak velocity of -0.5125 m/s, and then after the collision, it drops back down to 0.205 m/s for a bit and then eventually to 0.0 m/s (which it was 0.0 m/s at the initial start of the run as well). From my calculations above, I believe that the change in Kinetic energy is 0.0 J, which is a characteristic for an elastic collision because it means that the kinetic energy before and after the collision are the same.

v f v 1 i = m 1 m 1 + m 2 KE remote initial = ½ mv 2 = ½ (0.2047 kg)(0.02m/s) 2 = 0.00004094 = 4.1*10 -5 J KE cart initial = ½ mv 2 = ½ (0.3212 kg)(0.02kg) 2 = 0.00006424 J = 6.4*10 -5 J KE final = ½ (m 1 + m 2 ) v f 2 = ½ (0.2047+0.3212)(0.02) 2 = 0.00010518 = 1.1*10 -4 ΔKE initial - KE final –= (0.00004094+0.00006424) - 0.00010518 = 0 E lost = K E i − K E f = 1 2 m 1 v i 1 2 + 1 2 m 2 v i 2 2 − 1 2 ( m 1 + m 2 ) v f 2 The data indicates that this is an elastic collision because there is no kinetic energy lost. I’m not sure if maybe I calculated the wrong velocities because I was expecting there to be half energy gone. Part 1: Explosions Slide 10: Explosions Remote, m 1 = 204.1 g V max run 1 = 0.460 m/s V max run 2 = 0.540 m/s V max run 3 = 0.420 m/s K E run 1 = 1 2 mv 2 = ½ (204.1g)(0.460 m/s) 2 = 21.59378 K E run 2 = 1 2 m v 2 = ½ (204.1g)(0.540 m/s) 2 = 29.75778 K E run 3 = 1 2 m v 2 = ½ (204.1g)(0.420 m/s) 2 = 18.00162